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Proof step 6 from proof count-i2j-length (previous step)

i ≤ j, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → i0 ≤ j0 → # count-i2j(i0, j0) = i→n(j0 - i0) + 1 ⊦ # count-i2j(i + 1, j) = i→n(j - i)

User applied rule apply induction

And obtained the following new goals

Proof step 7:
i ≤ j, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → i0 ≤ j0 → # count-i2j(i0, j0) = i→n(j0 - i0) + 1 ⊦ i→n(j - i + 1) < i→n(j - i), # count-i2j(i + 1, j) = i→n(j - i)
Proof step 10:
# count-i2j(i + 1, j) = i→n(j - i + 1) + 1, i ≤ j, ∀ j0, i0. i→n(j0 - i0) < i→n(j - i) → i0 ≤ j0 → # count-i2j(i0, j0) = i→n(j0 - i0) + 1 ⊦ # count-i2j(i + 1, j) = i→n(j - i)

Rule argument

subst: [j0, i0] → [j, i + 1]

Additional information

Used simplifier rules: count-i2j-empty-02: ⊦ count-i2j(i, j) = [] ↔ j < i;%succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%sub-eq-01: ⊦ i = j → i - j = 0; from specification int-abs%l: ⊦ # ints = 0 ↔ ints = []; from specification list-data[intlista]%succ-02: ⊦ i < j + 1 ↔ i ≤ j; from specification int-abs Used terms: j, i + 1